# Proving a Line-integration along a parametrized curve identitiy.

(this question were asked after studying line integrals)

1- Show that if $$C$$ is the graph of $$y=f(x)$$, $$a \leq x \leq b,$$ and if $$F$$ is a function of 2 variables defined on C, then $$\int_{C} F(x,y)dx = \int_{a}^{b} F(x, f(x))dx.$$

But I do not understand the meaning of the question, does it means that if I change the graph by its parametrization I will change the integration along the graph to the integration at the end points of the interval $$[a,b]$$, if so why is this correct?

Also could anyone give me a hint for the proof please?

Thanks!

• Are you sure there is not a Jacobian missing, e.g. $\sqrt{1+f'(x)^2}$ ? – Thomas Apr 18 at 7:09
• @Thomas no there is not ..... this is exactly the question I have to answer. – Intuition Apr 18 at 7:11
• Looks a bit weird to me but maybe I am not understanding. The left member when F(x,y)=1 should not be the length of C ? When parametrizing the curve by $t \rightarrow [t,f(t)]$ tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx we would get a Jacobian more ? Maybe I am not getting the meaning of the let side... – Thomas Apr 18 at 7:19
• May be the question contains a typo (which I am not sure of it because my professor have seen the question and did not say anything about it ..... or may be he\she was not concentrating at that moment) ..... I do not know @Thomas – Intuition Apr 18 at 7:29
• Maybe I got the catch. I did not notice that you have a dx in the left member, which must be maybe intended as the integral of the differential form "F dx" over C. Before I interpreted the left integral as simply the integration of a scalar function over the manyfold C. – Thomas Apr 18 at 12:37

$$C$$ is defined by $$C := \{ (x,y) \ | \ x\in[a,b], \ f(x) = y \}$$ so $$\int_{C} F(x,y)dx = \int_{\substack{x\in [a,b] \\ y = f(x)}} F(x,y)dx = \int_{x\in[a,b]} F(x,f(x))dx =\int_a^b F(x,f(x))dx$$ This just represents any function along $$C$$. For example the length of the curve would be $$\int_C \sqrt{dx^2 + dy^2} = \int_C\sqrt{1 + (\frac{dy}{dx})^2}dx = \int_a^b \sqrt{1 + (f'(x))^2}dx$$ Here $$F(x,y) = \sqrt{1 + (\frac{dy}{dx})^2}$$

• what about the proof? – Intuition Apr 18 at 10:46
• I made an edit which should be enough justification. It could almost be used as a definition in my opinion. – Dayton Apr 18 at 10:56
• Thank you so much for you great effort. – Intuition Apr 18 at 10:59
• why is the first integration on the right hand side of the second line from below does not contain $dx$? – Smart Apr 18 at 13:46
• The infinitesimal Euclidean arc length is just $\sqrt{dx^2 + dy^2}$, which you can "factor" the $dx$ outside and have an integral depending only on one parameter. – Dayton Apr 18 at 13:56

Here one derivation starting from the definitions. Let $$\tau:[a,b]\rightarrow R^2:t\rightarrow(t,f(t))$$ a parametrization of $$C$$.

By definition the integral of a differential form $$\rho$$ is:

$$\int_C \rho =\int_{[a,b]} \rho(\tau(t))[\tau'(t)]dt$$

, where $$\rho(a)[b]$$ is the differential form at $$a$$ acting on vector $$b$$.

when $$\rho=Fdx$$, since $$\tau'(t)=(1,f'(t))$$, $$dx[\tau'(t)]=1$$ we have:

$$\int_C F(x,y) dx =\int_{[a,b]} F(t,f(t))dt$$

when $$\rho=Fdy$$, since $$\tau'(t)=(1,f'(t))$$, $$dy[\tau'(t)]=f'(t)$$ we have:

$$\int_C F(x,y) dy =\int_{[a,b]} F(t,f(t))f'(t)dt$$